Two burgers and one order of regular fries have 2020 calories. Four burgers and three orders of regular fries have 4660 calories. How many calories are there in each menu item?

There’s plenty not to like about this problem. As a real-world application of mathematics, like a Subway foot-long sandwich, it doesn’t measure up. For starters, it asks students to pretend that the total number of calories is known while the number of calories in each menu item remains a mystery. “We all use math everyday” meets “Yeah, right.” It would be easy to dismiss this problem. Pseudocontext. Full stop.

However, I also want to use the real world to have my students better understand mathematics. Whereas I’m critical of this problem presented as an application, I’m much more accepting of it as an investigation.

To introduce solving systems of linear equations, I have asked similar questions. My students would reason that if two burgers and one order of fries have 2020 calories, then four burgers and two orders of fries must have 2 times 2020, or 4040, calories. Comparing this with the number of calories in four burgers and three orders of fries means one extra order of fries adds 4660 – 4040, or 620, calories. Two burgers must then have 2020 – 620, or 1400, calories. Each burger has 1400 ÷ 2, or 700, calories. Students have solved a system of linear equations using the elimination method. All before having x‘s and y‘s thrown at them. My role was to help my students link their ideas within this context to this: